What is the slope of the line tangent to $f(x) = x^{2}-3x-5$ at $x = 3$ ?
Solution: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{((x+\Delta x)^{2}-3(x+\Delta x)-5) - (x^{2}-3x-5)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(x^{2}+2x \Delta x+\Delta x^{2}-3(x+\Delta x)-5) - (x^{2}-3x-5)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{x^{2}+2(x \Delta x)+\Delta x^{2}-3x-3(\Delta x)-5-x^{2}+3x+5}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{2(x \Delta x)+\Delta x^{2}-3(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} 2x+\Delta x-3$ $ = 2x-3$ $ = (2)(3)-3$ $ = 3$